All categories

1 year ago

7 pounds of potatoes cost $63. How many pounds ofpotatoes can you get with $261?

Mathematics

1 answer:

kozerog [31]1 year ago

3 0

We know that 7 pounds of potatoes cost $63.

This is a proportional relationship between quantity (pounds of potatoes) and cost.

We can calculate how many pounds of potatoes we can get with $261 applying the Rule of three:

$261\text{ \$}\cdot(\frac{7\text{ pounds}}{63\text{ \$}})=29\text{ pounds}$

We can buy 29 pounds of potatoes with $261

NOTE: I use the units ($ and pounds) to know how to multiply them in order to get the result I was looking for.

You might be interested in

Please help! 5 3/4-3 1/4=?

KonstantinChe [14]

Your answer will be 2 2/4. If you want it simplified it will be 2 1/2

6 0

3 years ago

Decide what information is needed to answer the following

oee [108]

Answer:

sgst

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

Step 3: Evaluate Functions

Substitute in center x [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center x [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center x [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center x [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

Step 4: Write Taylor Polynomial

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago

What is the quotient of x and three increased by 12 = 20

san4es73 [151]

Answer:

x= 24

Step-by-step explanation:

Quotient means divide, increased means add

x/3 + 12 =20

Subtract 12 from each side

x/3 +12-12 = 20-12

x/3=8

Multiply each side by 3

x/3 * 3 =8*3

x= 24

8 0

2 years ago

Read 2 more answers

Can someone please help me with this?

Contact [7]

Answer:

i need points

Step-by-step explanation:

i need points lol ,, hope you find your answer i guess .

4 0

3 years ago